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JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Question Bank Rate law and Rate Constant

  • question_answer
    In the reaction \[2{{N}_{2}}{{O}_{5}}\to 4N{{O}_{2}}+{{O}_{2}}\], initial pressure is \[500\,atm\] and rate constant \[K\] is \[3.38\times {{10}^{-5}}{{\sec }^{-1}}\]. After 10 minutes the final pressure of \[{{N}_{2}}{{O}_{5}}\] is               [Orissa JEE 2005]

    A)                 490 atm               

    B)                 250 atm

    C)                 480 atm               

    D)                 420 atm

    Correct Answer: A

    Solution :

               \[{{p}_{0}}=500\,atm\]                    \[K=\frac{2.303}{t}{{\log }_{10}}\frac{{{p}_{0}}}{{{p}_{t}}}\]                    \[3.38\times {{10}^{-5}}=\frac{2.303}{10\times 60}\log \frac{500}{{{p}_{t}}}\]                                 or \[0.00880=\log \frac{500}{{{p}_{t}}}\Rightarrow \frac{500}{1.02}=490\,atm\]


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